3.111 \(\int \frac {\sinh ^{-1}(a x)}{(c+a^2 c x^2)^{7/2}} \, dx\)

Optimal. Leaf size=200 \[ \frac {2}{15 a c^3 \sqrt {a^2 x^2+1} \sqrt {a^2 c x^2+c}}+\frac {1}{20 a c^3 \left (a^2 x^2+1\right )^{3/2} \sqrt {a^2 c x^2+c}}-\frac {4 \sqrt {a^2 x^2+1} \log \left (a^2 x^2+1\right )}{15 a c^3 \sqrt {a^2 c x^2+c}}+\frac {8 x \sinh ^{-1}(a x)}{15 c^3 \sqrt {a^2 c x^2+c}}+\frac {4 x \sinh ^{-1}(a x)}{15 c^2 \left (a^2 c x^2+c\right )^{3/2}}+\frac {x \sinh ^{-1}(a x)}{5 c \left (a^2 c x^2+c\right )^{5/2}} \]

[Out]

1/5*x*arcsinh(a*x)/c/(a^2*c*x^2+c)^(5/2)+4/15*x*arcsinh(a*x)/c^2/(a^2*c*x^2+c)^(3/2)+1/20/a/c^3/(a^2*x^2+1)^(3
/2)/(a^2*c*x^2+c)^(1/2)+8/15*x*arcsinh(a*x)/c^3/(a^2*c*x^2+c)^(1/2)+2/15/a/c^3/(a^2*x^2+1)^(1/2)/(a^2*c*x^2+c)
^(1/2)-4/15*ln(a^2*x^2+1)*(a^2*x^2+1)^(1/2)/a/c^3/(a^2*c*x^2+c)^(1/2)

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Rubi [A]  time = 0.12, antiderivative size = 200, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, Rules used = {5690, 5687, 260, 261} \[ \frac {2}{15 a c^3 \sqrt {a^2 x^2+1} \sqrt {a^2 c x^2+c}}+\frac {1}{20 a c^3 \left (a^2 x^2+1\right )^{3/2} \sqrt {a^2 c x^2+c}}-\frac {4 \sqrt {a^2 x^2+1} \log \left (a^2 x^2+1\right )}{15 a c^3 \sqrt {a^2 c x^2+c}}+\frac {8 x \sinh ^{-1}(a x)}{15 c^3 \sqrt {a^2 c x^2+c}}+\frac {4 x \sinh ^{-1}(a x)}{15 c^2 \left (a^2 c x^2+c\right )^{3/2}}+\frac {x \sinh ^{-1}(a x)}{5 c \left (a^2 c x^2+c\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[ArcSinh[a*x]/(c + a^2*c*x^2)^(7/2),x]

[Out]

1/(20*a*c^3*(1 + a^2*x^2)^(3/2)*Sqrt[c + a^2*c*x^2]) + 2/(15*a*c^3*Sqrt[1 + a^2*x^2]*Sqrt[c + a^2*c*x^2]) + (x
*ArcSinh[a*x])/(5*c*(c + a^2*c*x^2)^(5/2)) + (4*x*ArcSinh[a*x])/(15*c^2*(c + a^2*c*x^2)^(3/2)) + (8*x*ArcSinh[
a*x])/(15*c^3*Sqrt[c + a^2*c*x^2]) - (4*Sqrt[1 + a^2*x^2]*Log[1 + a^2*x^2])/(15*a*c^3*Sqrt[c + a^2*c*x^2])

Rule 260

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rule 261

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a + b*x^n)^(p + 1)/(b*n*(p + 1)), x] /; FreeQ
[{a, b, m, n, p}, x] && EqQ[m, n - 1] && NeQ[p, -1]

Rule 5687

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)/((d_) + (e_.)*(x_)^2)^(3/2), x_Symbol] :> Simp[(x*(a + b*ArcSinh
[c*x])^n)/(d*Sqrt[d + e*x^2]), x] - Dist[(b*c*n*Sqrt[1 + c^2*x^2])/(d*Sqrt[d + e*x^2]), Int[(x*(a + b*ArcSinh[
c*x])^(n - 1))/(1 + c^2*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d] && GtQ[n, 0]

Rule 5690

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[(x*(d + e*x^2)^(p
+ 1)*(a + b*ArcSinh[c*x])^n)/(2*d*(p + 1)), x] + (Dist[(2*p + 3)/(2*d*(p + 1)), Int[(d + e*x^2)^(p + 1)*(a +
b*ArcSinh[c*x])^n, x], x] + Dist[(b*c*n*d^IntPart[p]*(d + e*x^2)^FracPart[p])/(2*(p + 1)*(1 + c^2*x^2)^FracPar
t[p]), Int[x*(1 + c^2*x^2)^(p + 1/2)*(a + b*ArcSinh[c*x])^(n - 1), x], x]) /; FreeQ[{a, b, c, d, e}, x] && EqQ
[e, c^2*d] && GtQ[n, 0] && LtQ[p, -1] && NeQ[p, -3/2]

Rubi steps

\begin {align*} \int \frac {\sinh ^{-1}(a x)}{\left (c+a^2 c x^2\right )^{7/2}} \, dx &=\frac {x \sinh ^{-1}(a x)}{5 c \left (c+a^2 c x^2\right )^{5/2}}+\frac {4 \int \frac {\sinh ^{-1}(a x)}{\left (c+a^2 c x^2\right )^{5/2}} \, dx}{5 c}-\frac {\left (a \sqrt {1+a^2 x^2}\right ) \int \frac {x}{\left (1+a^2 x^2\right )^3} \, dx}{5 c^3 \sqrt {c+a^2 c x^2}}\\ &=\frac {1}{20 a c^3 \left (1+a^2 x^2\right )^{3/2} \sqrt {c+a^2 c x^2}}+\frac {x \sinh ^{-1}(a x)}{5 c \left (c+a^2 c x^2\right )^{5/2}}+\frac {4 x \sinh ^{-1}(a x)}{15 c^2 \left (c+a^2 c x^2\right )^{3/2}}+\frac {8 \int \frac {\sinh ^{-1}(a x)}{\left (c+a^2 c x^2\right )^{3/2}} \, dx}{15 c^2}-\frac {\left (4 a \sqrt {1+a^2 x^2}\right ) \int \frac {x}{\left (1+a^2 x^2\right )^2} \, dx}{15 c^3 \sqrt {c+a^2 c x^2}}\\ &=\frac {1}{20 a c^3 \left (1+a^2 x^2\right )^{3/2} \sqrt {c+a^2 c x^2}}+\frac {2}{15 a c^3 \sqrt {1+a^2 x^2} \sqrt {c+a^2 c x^2}}+\frac {x \sinh ^{-1}(a x)}{5 c \left (c+a^2 c x^2\right )^{5/2}}+\frac {4 x \sinh ^{-1}(a x)}{15 c^2 \left (c+a^2 c x^2\right )^{3/2}}+\frac {8 x \sinh ^{-1}(a x)}{15 c^3 \sqrt {c+a^2 c x^2}}-\frac {\left (8 a \sqrt {1+a^2 x^2}\right ) \int \frac {x}{1+a^2 x^2} \, dx}{15 c^3 \sqrt {c+a^2 c x^2}}\\ &=\frac {1}{20 a c^3 \left (1+a^2 x^2\right )^{3/2} \sqrt {c+a^2 c x^2}}+\frac {2}{15 a c^3 \sqrt {1+a^2 x^2} \sqrt {c+a^2 c x^2}}+\frac {x \sinh ^{-1}(a x)}{5 c \left (c+a^2 c x^2\right )^{5/2}}+\frac {4 x \sinh ^{-1}(a x)}{15 c^2 \left (c+a^2 c x^2\right )^{3/2}}+\frac {8 x \sinh ^{-1}(a x)}{15 c^3 \sqrt {c+a^2 c x^2}}-\frac {4 \sqrt {1+a^2 x^2} \log \left (1+a^2 x^2\right )}{15 a c^3 \sqrt {c+a^2 c x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.18, size = 121, normalized size = 0.60 \[ \frac {\sqrt {a^2 c x^2+c} \left (4 a x \sqrt {a^2 x^2+1} \left (8 a^4 x^4+20 a^2 x^2+15\right ) \sinh ^{-1}(a x)-\left (a^2 x^2+1\right ) \left (-8 a^2 x^2+16 \left (a^2 x^2+1\right )^2 \log \left (a^2 x^2+1\right )-11\right )\right )}{60 a c^4 \left (a^2 x^2+1\right )^{7/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcSinh[a*x]/(c + a^2*c*x^2)^(7/2),x]

[Out]

(Sqrt[c + a^2*c*x^2]*(4*a*x*Sqrt[1 + a^2*x^2]*(15 + 20*a^2*x^2 + 8*a^4*x^4)*ArcSinh[a*x] - (1 + a^2*x^2)*(-11
- 8*a^2*x^2 + 16*(1 + a^2*x^2)^2*Log[1 + a^2*x^2])))/(60*a*c^4*(1 + a^2*x^2)^(7/2))

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fricas [F]  time = 0.59, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {a^{2} c x^{2} + c} \operatorname {arsinh}\left (a x\right )}{a^{8} c^{4} x^{8} + 4 \, a^{6} c^{4} x^{6} + 6 \, a^{4} c^{4} x^{4} + 4 \, a^{2} c^{4} x^{2} + c^{4}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsinh(a*x)/(a^2*c*x^2+c)^(7/2),x, algorithm="fricas")

[Out]

integral(sqrt(a^2*c*x^2 + c)*arcsinh(a*x)/(a^8*c^4*x^8 + 4*a^6*c^4*x^6 + 6*a^4*c^4*x^4 + 4*a^2*c^4*x^2 + c^4),
x)

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giac [A]  time = 0.39, size = 124, normalized size = 0.62 \[ -\frac {1}{60} \, \sqrt {c} {\left (\frac {16 \, \log \left (a^{2} x^{2} + 1\right )}{a c^{4}} - \frac {24 \, a^{4} x^{4} + 56 \, a^{2} x^{2} + 35}{{\left (a^{2} x^{2} + 1\right )}^{2} a c^{4}}\right )} + \frac {{\left (4 \, {\left (\frac {2 \, a^{4} x^{2}}{c} + \frac {5 \, a^{2}}{c}\right )} x^{2} + \frac {15}{c}\right )} x \log \left (a x + \sqrt {a^{2} x^{2} + 1}\right )}{15 \, {\left (a^{2} c x^{2} + c\right )}^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsinh(a*x)/(a^2*c*x^2+c)^(7/2),x, algorithm="giac")

[Out]

-1/60*sqrt(c)*(16*log(a^2*x^2 + 1)/(a*c^4) - (24*a^4*x^4 + 56*a^2*x^2 + 35)/((a^2*x^2 + 1)^2*a*c^4)) + 1/15*(4
*(2*a^4*x^2/c + 5*a^2/c)*x^2 + 15/c)*x*log(a*x + sqrt(a^2*x^2 + 1))/(a^2*c*x^2 + c)^(5/2)

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maple [B]  time = 0.22, size = 363, normalized size = 1.82 \[ \frac {16 \sqrt {c \left (a^{2} x^{2}+1\right )}\, \arcsinh \left (a x \right )}{15 \sqrt {a^{2} x^{2}+1}\, a \,c^{4}}+\frac {\sqrt {c \left (a^{2} x^{2}+1\right )}\, \left (8 x^{5} a^{5}-8 \sqrt {a^{2} x^{2}+1}\, x^{4} a^{4}+20 x^{3} a^{3}-16 \sqrt {a^{2} x^{2}+1}\, x^{2} a^{2}+15 a x -8 \sqrt {a^{2} x^{2}+1}\right ) \left (-64 x^{8} a^{8}-64 \sqrt {a^{2} x^{2}+1}\, x^{7} a^{7}-280 x^{6} a^{6}-248 \sqrt {a^{2} x^{2}+1}\, x^{5} a^{5}+160 \arcsinh \left (a x \right ) x^{4} a^{4}-456 x^{4} a^{4}-340 \sqrt {a^{2} x^{2}+1}\, x^{3} a^{3}+380 \arcsinh \left (a x \right ) x^{2} a^{2}-328 a^{2} x^{2}-165 \sqrt {a^{2} x^{2}+1}\, x a +256 \arcsinh \left (a x \right )-88\right )}{60 \left (40 a^{10} x^{10}+215 x^{8} a^{8}+469 x^{6} a^{6}+517 x^{4} a^{4}+287 a^{2} x^{2}+64\right ) a \,c^{4}}-\frac {8 \sqrt {c \left (a^{2} x^{2}+1\right )}\, \ln \left (1+\left (a x +\sqrt {a^{2} x^{2}+1}\right )^{2}\right )}{15 \sqrt {a^{2} x^{2}+1}\, a \,c^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arcsinh(a*x)/(a^2*c*x^2+c)^(7/2),x)

[Out]

16/15*(c*(a^2*x^2+1))^(1/2)/(a^2*x^2+1)^(1/2)/a/c^4*arcsinh(a*x)+1/60*(c*(a^2*x^2+1))^(1/2)*(8*x^5*a^5-8*(a^2*
x^2+1)^(1/2)*x^4*a^4+20*x^3*a^3-16*(a^2*x^2+1)^(1/2)*x^2*a^2+15*a*x-8*(a^2*x^2+1)^(1/2))*(-64*x^8*a^8-64*(a^2*
x^2+1)^(1/2)*x^7*a^7-280*x^6*a^6-248*(a^2*x^2+1)^(1/2)*x^5*a^5+160*arcsinh(a*x)*x^4*a^4-456*x^4*a^4-340*(a^2*x
^2+1)^(1/2)*x^3*a^3+380*arcsinh(a*x)*x^2*a^2-328*a^2*x^2-165*(a^2*x^2+1)^(1/2)*x*a+256*arcsinh(a*x)-88)/(40*a^
10*x^10+215*a^8*x^8+469*a^6*x^6+517*a^4*x^4+287*a^2*x^2+64)/a/c^4-8/15*(c*(a^2*x^2+1))^(1/2)/(a^2*x^2+1)^(1/2)
/a/c^4*ln(1+(a*x+(a^2*x^2+1)^(1/2))^2)

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maxima [A]  time = 0.41, size = 143, normalized size = 0.72 \[ \frac {1}{60} \, a {\left (\frac {3}{{\left (a^{6} c^{\frac {5}{2}} x^{4} + 2 \, a^{4} c^{\frac {5}{2}} x^{2} + a^{2} c^{\frac {5}{2}}\right )} c} + \frac {8}{{\left (a^{4} c^{\frac {3}{2}} x^{2} + a^{2} c^{\frac {3}{2}}\right )} c^{2}} - \frac {16 \, \log \left (x^{2} + \frac {1}{a^{2}}\right )}{a^{2} c^{\frac {7}{2}}}\right )} + \frac {1}{15} \, {\left (\frac {8 \, x}{\sqrt {a^{2} c x^{2} + c} c^{3}} + \frac {4 \, x}{{\left (a^{2} c x^{2} + c\right )}^{\frac {3}{2}} c^{2}} + \frac {3 \, x}{{\left (a^{2} c x^{2} + c\right )}^{\frac {5}{2}} c}\right )} \operatorname {arsinh}\left (a x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsinh(a*x)/(a^2*c*x^2+c)^(7/2),x, algorithm="maxima")

[Out]

1/60*a*(3/((a^6*c^(5/2)*x^4 + 2*a^4*c^(5/2)*x^2 + a^2*c^(5/2))*c) + 8/((a^4*c^(3/2)*x^2 + a^2*c^(3/2))*c^2) -
16*log(x^2 + 1/a^2)/(a^2*c^(7/2))) + 1/15*(8*x/(sqrt(a^2*c*x^2 + c)*c^3) + 4*x/((a^2*c*x^2 + c)^(3/2)*c^2) + 3
*x/((a^2*c*x^2 + c)^(5/2)*c))*arcsinh(a*x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\mathrm {asinh}\left (a\,x\right )}{{\left (c\,a^2\,x^2+c\right )}^{7/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(asinh(a*x)/(c + a^2*c*x^2)^(7/2),x)

[Out]

int(asinh(a*x)/(c + a^2*c*x^2)^(7/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {asinh}{\left (a x \right )}}{\left (c \left (a^{2} x^{2} + 1\right )\right )^{\frac {7}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(asinh(a*x)/(a**2*c*x**2+c)**(7/2),x)

[Out]

Integral(asinh(a*x)/(c*(a**2*x**2 + 1))**(7/2), x)

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